# Why is lattice QCD called non-perturbative?

Like, if you are approximating a smooth structure with a discrete lattice, isn't this like a perturbation from smooth space-time?

If Feynman diagrams are a perturbative method, why are Feynamn diagrams on a lattice/grid called non-perturbative?

• My mistake, I assumed lattice QCD was the same as QCD Feynman diagrams on a finite grid. – zooby Dec 22 '18 at 21:35

In general, by a perturbative approach, we mean an approximation of the form,

$$f = f_0 + \epsilon f_1 + \epsilon^2 f_2 + \dots$$

where $$\epsilon$$ is the perturbation parameter, for some solution $$f$$. That is to say, one can approximate the behaviour of the solution by this series.

However, summing all the terms does not mean you will recover the exact solution; in most cases perturbative series are asymptotic series.

On the other hand, lattice QCD is an approach which is not described as perturbation theory because it does not follow this scheme, and in principle one recovers the exact solution in the appropriate limit.

To convince yourself of the distinction, consider the differential equation,

$$\frac{\mathrm d f}{\mathrm dx} = g(x).$$

If we choose to discretize it (very naively), we obtain a linear system,

$$\frac{f_{i+1}-f_{i}}{\Delta x} = g_i$$

which is a totally different approach to plugging in a series expansion like the one above. That being said, Feynman diagrams are always a perturbative approach, so Feynman diagrams on a lattice are a perturbative approach, but putting the theory on a lattice is not what makes it perturbative.

• So in general it's just the kind of limit your taking. – zooby Dec 22 '18 at 21:33
• @zooby: For a perturbative expansion, the limit does not exist. (This is what "asymptotic series" means.) For lattice QCD, as far as we can tell, the limit exists. – Peter Shor Dec 22 '18 at 22:42
• @Peter do you know if string theory perturbation theory uses asymptotic expansions? – zooby Dec 23 '18 at 0:03
• @zooby String perturbation theory replaces Feynman diagrams with a higher dimensional analogue, namely Riemann surfaces. It's still a perturbative expansion. – JamalS Dec 23 '18 at 21:47
• zooby didn't ask whether string perturbation theory is perturbative (which is kind of obvious from the name), but whether it's asymptotic. – tparker Dec 25 '18 at 18:07